2 Eggs and A Building

[Osprey]

They can break, they might not
Building is tall, having hundred floors
Tell the times, should we throw
The wobbly and brittly eggs.
So that , we reach the highest place.

[Hàkon]

In the worst case scenario we have to drop the eggs 99 times, that is also with only one egg and no possibility to guess one floor too high.

In that case we need to start at the bottom floor.

But with two eggs, and thus the possibility to guess one floor too high we can start anywhere.

We can start at any floor F, if the egg breaks the next tests wil be done on the floors below, if not we move upwards.

Anyway, now we have F-1 remainning drops, and the logic above is applied: we move upwards if the egg does not break, F + (F-1).

In time we move up as high as it goes: F + (F-1) + (F-2) + ... + 1 <= 100

This is a quadratic equation that gives the smallest value of F when F(F-1)/ 2 ≥ 99 to be 14.

The least amount of throws is 14.

[Corvus]

In the worst case scenario we have to drop the eggs 99 times, that is also with only one egg and no possibility to guess one floor too high.

In that case we need to start at the bottom floor.

But with two eggs, and thus the possibility to guess one floor too high we can start anywhere.

We can start at any floor F, if the egg breaks the next tests wil be done on the floors below, if not we move upwards.

Anyway, now we have F-1 remainning drops, and the logic above is applied: we move upwards if the egg does not break, F + (F-1).

In time we move up as high as it goes: F + (F-1) + (F-2) + ... + 1 <= 100

This is a quadratic equation that gives the smallest value of F when F(F-1)/ 2 ≥ 99 to be 14.

The least amounts of throws are 14.

Test passed
You are hired :D

[Osprey]

Too easy for Apricians!
Let me think up some tough one!

[Osprey]

Test passed
You are hired :D

For the Corvus Massage Club, women only!:D

[Hàkon]

I've thought out one, if you're interested that is.

[Hàkon]

Test passed
I'll massage you.


For the Corvus Massage Club, women only!:D

I'd like a massage.

[Osprey]

I'd like a massage.

Take care, you might fall in love with the muscled, romantic massager........

[Quasimodem]

In the worst case scenario we have to drop the eggs 99 times, that is also with only one egg and no possibility to guess one floor too high.

In that case we need to start at the bottom floor.

But with two eggs, and thus the possibility to guess one floor too high we can start anywhere.

We can start at any floor F, if the egg breaks the next tests wil be done on the floors below, if not we move upwards.

Anyway, now we have F-1 remainning drops, and the logic above is applied: we move upwards if the egg does not break, F + (F-1).

In time we move up as high as it goes: F + (F-1) + (F-2) + ... + 1 <= 100

This is a quadratic equation that gives the smallest value of F when F(F-1)/ 2 ≥ 99 to be 14.

The least amount of throws is 14.

I think I'm in love.

[Dacul]

I'd like a massage.

I can give you a message.
Is that fine instead of a massage?

[Hàkon]

I can give you a message.
Is that fine instead of a massage?

All is welcomed.

[Hàkon]

Well, since Osprey has not provided a new nut to crack I have come up with my own:


I have three coins in my pocket that I, with huge precision, fit into the smallest equilateral triangle conceivable.

All three coins fit exactly, and sit as close next to each other as possible.

How great is the percentage of the triangle's area which is not covered by the coins?

[el22]

If I've understood it right (and haven't done any errors) the solution should go:
r = coin radius
Triangle area = r*r*(tan(30)*4 + 6)
coin area = 3*pi*r*r
no calculator to calculate the actual value :(

[Osprey]

Well, since Osprey has not provided a new nut to crack I have come up with my own:


I have three coins in my pocket that I, with huge precision, fit into the smallest equilateral triangle conceivable.

All three coins fit exactly, and sit as close next to each other as possible.

How great is the percentage of the triangle's area which is not covered by the coins?

Sorry, couldn't get any puzzle which might puzzle you :)
The answer is 16.something percent, if i have my calculations right.
Here's another one. (pretty simple)
I have 100 coins on the table. Out of them exactly 50 are heads up. You are blindfolded. You have to seperate the coins into two stacks with equal number of heads up. And you can't feel the coins. (assume a paper covers each side of the coin)

[Quasimodem]

Well, since Osprey has not provided a new nut to crack I have come up with my own:


I have three coins in my pocket that I, with huge precision, fit into the smallest equilateral triangle conceivable.

All three coins fit exactly, and sit as close next to each other as possible.

How great is the percentage of the triangle's area which is not covered by the coins?

Okay, is it 21.46%?

[Osprey]

Is it...

http://imageshack.us/a/img15/99/cointriangleanswer.png????

What a cute, childlike handwriting?!!

[el22]

I've done an error earlier on, tan(30) should be 1/tan(30) and if everything else is OK the answer should be 39%

[el22]

Sorry, couldn't get any puzzle which might puzzle you :)
The answer is 16.something percent, if i have my calculations right.
Here's another one. (pretty simple)
I have 100 coins on the table. Out of them exactly 50 are heads up. You are blindfolded. You have to seperate the coins into two stacks with equal number of heads up. And you can't feel the coins. (assume a paper covers each side of the coin)

My brother told me about a trick with cards he had seen in youtube, similar to this problem. So the solution should be:
1. divide the coins in two groups with equal number of coins
2. turn over the coins of one group

[Quasimodem]

What a cute, childlike handwriting?!!

Well, it's a true Microsoft Paint Masterpiece. Art at its finest. But now I'm getting 21.46%...

[Hàkon]

I've done an error earlier on, tan(30) should be 1/tan(30) and if everything else is OK the answer should be 39%


What a cute, childlike handwriting?!!


Okay, is it 21.46%?

All wrong.

Also, I would like to see your full calculations.

Would you like to continue guessing or do you perhaps want the answer?

[el22]

OK, my calculations:
triangle area=
8*r*r/(2*tan(30)) + 6*r*r = 2*r*r*(1/tan(30) + 3)
coin area = 3*pi*r*r

% = 100 * (1 - (3*pi)/(2(2/(tan(30) + 3)) = 50%

[Quasimodem]

All wrong.

Also, I would like to see your full calculations.

Would you like to continue guessing or do you perhaps want the answer?

After fixing a dumb mistake...~27%???

Don't give us the answer, by the way. And I'll show my calculations once/if I get it right!

EDIT: Are all these coins the same size?

[el22]

= 27.1% now
I give up

Strmstn can I go now? :)

[Hàkon]

After fixing a dumb mistake...~27%???

Don't give us the answer, by the way. And I'll show my calculations once/if I get it right!

≈ 27% is right!

My calculation:

r = the radius of the circles

The uniformity principle gives the distance from origo in a circle to the nearest triangle tip = 2r.

The third side of the triangle that is formed by the radius in a circle and origo to the triangle tip = sqr(4*r^2-r^2) = r*sqr(3)

Side in the outer triangle = 2*r*sqr(3)+2*r = 2*r(sqr(3)+1)

Height of the outer triangle = 2*r+sqr(3)*r+r = r*(sqr(3)+3)

Area of the triangle TA (YES, TA!) = 2*r*(sqr(3)+1)*r*(sqr(3)+3)/2 =
r^2*(6+4*sqr(3))

Total area of the circles CA: 3*pi*r^2

Area not covered by the circles = (r^2*(6+4*sqr(3))-3*pi*r^2)/(r^2*(6+4*sqr(3)))*100 = (6+4*sqr(3)-3*pi)/(6+4*sqr(3))*100 ≈ 27,1%

[el22]

StrmStn I gave you a more simple solution, without even being aware of the 'uniformity principle' (whatever it means), and (apart from a calculation error) my answer was right

[Hàkon]

StrmStn I gave you a more simple solution, without even being aware of the 'uniformity principle' (whatever it means), and (apart from a calculation error) my answer was right

I actually did not see the first post you made, I apologize for that.

I searched the thread hastily by Ctrl+F + %.

[Osprey]

keep the calculations to a minimum please...

[Quasimodem]

≈ 27% is right!

My calculation:

r = the radius of the circles

The uniformity principle gives the distance from origo in a circle to the nearest triangle tip = 2r.

The third side of the triangle that is formed by the radius in a circle and origo to the triangle tip = sqr(4*r^2-r^2) = r*sqr(3)

Side in the outer triangle = 2*r*sqr(3)+2*r = 2*r(sqr(3)+1)

Height of the outer triangle = 2*r+sqr(3)*r+r = r*(sqr(3)+3)

Area of the triangle TA (YES, TA!) = 2*r*(sqr(3)+1)*r*(sqr(3)+3)/2 =
r^2*(6+4*sqr(3))

Total area of the circles CA: 3*pi*r^2

Area not covered by the circles = (r^2*(6+4*sqr(3))-3*pi*r^2)/(r^2*(6+4*sqr(3)))*100 = (6+4*sqr(3)-3*pi)/(6+4*sqr(3))*100 ≈ 27,1%

Here's how I did it (unfortunately, some was cut off while scanning, but you can probably get the gist of it):

http://imageshack.us/a/img819/7194/cointriangleproblem.jpg